How to find the limit of the sequence given by $X_{n+1} = \frac12 X_n + \frac1{X_n}$ $X_0 := 2$
and for $X_n$:
$X_{n+1} = \frac12 X_n + \frac1{X_n}$
I know that the sequence is monotone decreasing and I am not sure how to find its limit maybe with Cauchy and partial sums but still I cannot find the exact value of $\lim X_n$ for $n \rightarrow \infty$
 A: Hint:
$x_{n+1}$ is a subsequence of $x_n$, so they have the same limit as $n \rightarrow \infty$, provided the limit exists.  You already know the sequence is monotone decreasing, so now you'll want to show that it is also bounded below.  Once you have that, then a limit $L \neq \pm \infty$ exists by the monotone convergence theorem.  At this point, you can simply take a limit of both sides and apply limit arithmetic:
$$\lim(x_{n+1}) = \lim \left( \frac{x_n}{2} + \frac{1}{x_n} \right) = \frac{\lim(x_n)}{2} + \frac{1}{\lim(x_n)}$$
In other words:
$$L = \frac{L}{2} + \frac{1}{L}$$
And now it's just a matter of solving for $L$.
A: $$\begin{array}{l}
 \mathop {\lim }\limits_{n \to  + \infty } x_{n + 1}  = \mathop {\lim }\limits_{n \to  + \infty } x_n  = \ell  \\ 
  \Leftrightarrow \ell  = \frac{\ell }{2} + \frac{1}{\ell } \\ 
  \Leftrightarrow \ell  = \frac{{\ell ^2 }}{{2\ell }} + \frac{2}{{2\ell }} \\ 
  \Leftrightarrow 2\ell ^2  - \ell ^2  - 2 = 0 \\ 
  \Leftrightarrow \ell  \in \left\{ {\frac{{1 - \sqrt {17} }}{4};\frac{{1 + \sqrt {17} }}{4}} \right\} \\ 
 \end{array}$$
but the positive result that is to say,
$${\ell  = \frac{{1 + \sqrt {17} }}{4}}$$
finally$$
{\mathop {\lim }\limits_{n \to  + \infty } x_n  = \frac{{1 + \sqrt {17} }}{4}}
$$
